## Mahalanobis Distance using Eigen-Values of the Covariance Matrix

Given the formula of Mahalanobis Distance:

$D^2_M = (\mathbf{x} - \mathbf{\mu})^T \mathbf{S}^{-1} (\mathbf{x} - \mathbf{\mu})$

If I simplify the above expression using Eigen-value decomposition (EVD) of the Covariance Matrix:

$S = \mathbf{P} \Lambda \mathbf{P}^T$

Then,

$D^2_M = (\mathbf{x} - \mathbf{\mu})^T \mathbf{P} \Lambda^{-1} \mathbf{P}^T (\mathbf{x} - \mathbf{\mu})$

Let, the projections of $(\mathbf{x}-\mu)$ on all eigen-vectors present in $\mathbf{P}$ be $\mathbf{b}$, then:

$\mathbf{b} = \mathbf{P}^T(\mathbf{x} - \mathbf{\mu})$

And,

$D^2_M = \mathbf{b}^T \Lambda^{-1} \mathbf{b}$

$D^2_M = \sum_i{\frac{b^2_i}{\lambda_i}}$

The problem that I am facing right now is as follows:

The covariance matrix $\mathbf{S}$ is calculated on a dataset, in which no. of observations are less than the no. of variables. This causes some zero-valued eigen-values after EVD of $\mathbf{S}$.

In these cases the above simplified expression does not result in the same Mahalanobis Distance as the original expression, i.e.:

$(\mathbf{x} - \mathbf{\mu})^T \mathbf{S}^{-1} (\mathbf{x} - \mathbf{\mu}) \neq \sum_i{\frac{b^2_i}{\lambda_i}}$ (for non-zero $\lambda_i$)

My question is: Is the simplified expression still functionally represents the Mahalanobis Distance?

P.S.: Motivation to use the simplified expression of Mahalanbis Distance is to calculate its gradient wrt $b$.
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 Hello, In order to be invertible, S mustn't have zero eigen values, that is , must be positive definite or negative definite. Apart from that , that expression must work... All the best GoodSpirit
 Hey orajput and welcome to the forums. For your problem, if you do have a singular or ill-conditioned covariance matrix, I would try and do something like Principal Components, or to remove the offending variable from your system and re-do the analysis.

 Tags eigenvalues, eigenvectors, linear-algebra, mahalanobis